Homework Answers
1) 2 mL * 45 g/100 mL = 0.9 g
So, there is 0.9 g of ammonium sulfate in 2 mL of 45% corresponding solution.
2) M1V1 = M2V2
i.e. 1 M * V1 = 20*10-3 M * 50 mL
i.e. The volume of stock solution required (V1) = 1 mL
3) Moles of glycine = (50/1000) L * 1 mol/L = 0.05 mol
i.e. The mass of glycine required = 0.05 mol * 75.07 g/mol = 3.7535 g
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